Linear Algebra/Strings/Solutions

SolutionsEdit

This exercise is recommended for all readers.
Problem 1

What is the index of nilpotency of the left-shift operator, here acting on the space of triples of reals?

 
Answer

Three. It is at least three because  . It is at most three because  .

This exercise is recommended for all readers.
Problem 2

For each string basis state the index of nilpotency and give the dimension of the rangespace and nullspace of each iteration of the nilpotent map.

  1.  
  2.  
  3.  

Also give the canonical form of the matrix.

Answer
  1. The domain has dimension four. The map's action is that any vector in the space   is sent to  . The first application of the map sends two basis vectors   and   to zero, and therefore the nullspace has dimension two and the rangespace has dimension two. With a second application, all four basis vectors are sent to zero and so the nullspace of the second power has dimension four while the rangespace of the second power has dimension zero. Thus the index of nilpotency is two. This is the canonical form.
     
  2. The dimension of the domain of this map is six. For the first power the dimension of the nullspace is four and the dimension of the rangespace is two. For the second power the dimension of the nullspace is five and the dimension of the rangespace is one. Then the third iteration results in a nullspace of dimension six and a rangespace of dimension zero. The index of nilpotency is three, and this is the canonical form.
     
  3. The dimension of the domain is three, and the index of nilpotency is three. The first power's null space has dimension one and its range space has dimension two. The second power's null space has dimension two and its range space has dimension one. Finally, the third power's null space has dimension three and its range space has dimension zero. Here is the canonical form matrix.
     
Problem 3

Decide which of these matrices are nilpotent.

  1.  
  2.  
  3.  
  4.  
  5.  
Answer

By Lemma 1.3 the nullity has grown as large as possible by the  -th iteration where   is the dimension of the domain. Thus, for the   matrices, we need only check whether the square is the zero matrix. For the   matrices, we need only check the cube.

  1. Yes, this matrix is nilpotent because its square is the zero matrix.
  2. No, the square is not the zero matrix.
     
  3. Yes, the cube is the zero matrix. In fact, the square is zero.
  4. No, the third power is not the zero matrix.
     
  5. Yes, the cube of this matrix is the zero matrix.

Another way to see that the second and fourth matrices are not nilpotent is to note that they are nonsingular.

This exercise is recommended for all readers.
Problem 4

Find the canonical form of this matrix.

 
Answer

The table os calculations

 

gives these requirements of the string basis: three basis vectors are sent directly to zero, one more basis vector is sent to zero by a second application, and the final basis vector is sent to zero by a third application. Thus, the string basis has this form.

 

From that the canonical form is immediate.

 
This exercise is recommended for all readers.
Problem 5

Consider the matrix from Example 2.16.

  1. Use the action of the map on the string basis to give the canonical form.
  2. Find the change of basis matrices that bring the matrix to canonical form.
  3. Use the answer in the prior item to check the answer in the first item.
Answer
  1. The canonical form has a   block and a   block
     
    corresponding to the length three string and the length two string in the basis.
  2. Assume that   is the representation of the underlying map with respect to the standard basis. Let   be the basis to which we will change. By the similarity diagram
     
    we have that the canonical form matrix is   where
     
    and   is the inverse of that.
     
  3. The calculation to check this is routine.
This exercise is recommended for all readers.
Problem 6

Each of these matrices is nilpotent.

  1.  
  2.  
  3.  

Put each in canonical form.

Answer
  1. The calculation
     

    shows that any map represented by the matrix must act on the string basis in this way

     

    because the nullspace after one application has dimension one and exactly one basis vector,  , is sent to zero. Therefore, this representation with respect to   is the canonical form.

     
  2. The calculation here is similar to the prior one.
     

    The table shows that the string basis is of the form

     

    because the nullspace after one application of the map has dimension two—   and   are both sent to zero— and one more iteration results in the additional vector being brought to zero.

  3. The calculation
     

    shows that any map represented by this basis must act on a string basis in this way.

     

    Therefore, this is the canonical form.

     
Problem 7

Describe the effect of left or right multiplication by a matrix that is in the canonical form for nilpotent matrices.

Answer

A couple of examples

 

suggest that left multiplication by a block of subdiagonal ones shifts the rows of a matrix downward. Distinct blocks

 

act to shift down distinct parts of the matrix.

Right multiplication does an analgous thing to columns. See Problem 1.

Problem 8

Is nilpotence invariant under similarity? That is, must a matrix similar to a nilpotent matrix also be nilpotent? If so, with the same index?

Answer

Yes. Generalize the last sentence in Example 2.9. As to the index, that same last sentence shows that the index of the new matrix is less than or equal to the index of  , and reversing the roles of the two matrices gives inequality in the other direction.

Another answer to this question is to show that a matrix is nilpotent if and only if any associated map is nilpotent, and with the same index. Then, because similar matrices represent the same map, the conclusion follows. This is Exercise 14 below.

This exercise is recommended for all readers.
Problem 9

Show that the only eigenvalue of a nilpotent matrix is zero.

Answer

Observe that a canonical form nilpotent matrix has only zero eigenvalues; e.g., the determinant of this lower-triangular matrix

 

is  , the only root of which is zero. But similar matrices have the same eigenvalues and every nilpotent matrix is similar to one in canonical form.

Another way to see this is to observe that a nilpotent matrix sends all vectors to zero after some number of iterations, but that conflicts with an action on an eigenspace   unless   is zero.

Problem 10

Is there a nilpotent transformation of index three on a two-dimensional space?

Answer

No, by Lemma 1.3 for a map on a two-dimensional space, the nullity has grown as large as possible by the second iteration.

Problem 11

In the proof of Theorem 2.13, why isn't the proof's base case that the index of nilpotency is zero?

Answer

The index of nilpotency of a transformation can be zero only when the vector starting the string must be  , that is, only when   is a trivial space.

This exercise is recommended for all readers.
Problem 12

Let   be a linear transformation and suppose   is such that   but  . Consider the  -string  .

  1. Prove that   is a transformation on the span of the set of vectors in the string, that is, prove that   restricted to the span has a range that is a subset of the span. We say that the span is a  -invariant subspace.
  2. Prove that the restriction is nilpotent.
  3. Prove that the  -string is linearly independent and so is a basis for its span.
  4. Represent the restriction map with respect to the  -string basis.
Answer
  1. Any member   of the span can be written as a linear combination  . But then, by the linearity of the map,   is also in the span.
  2. The operation in the prior item, when iterated   times, will result in a linear combination of zeros.
  3. If   then the set is empty and so is linearly independent by definition. Otherwise write   and apply   to both sides. The right side gives   while the left side gives  ; conclude that  . Continue in this way by applying   to both sides, etc.
  4. Of course,   acts on the span by acting on this basis as a single,  -long,  -string.
     
Problem 13

Finish the proof of Theorem 2.13.

Answer

We must check that   is linearly independent where   is a  -string basis for  , where   is a basis for  , and where  . Write

 

and apply  .

 

Conclude that the coefficients   are all zero as   is a basis. Substitute back into the first displayed equation to conclude that the remaining coefficients are zero also.

Problem 14

Show that the terms "nilpotent transformation" and "nilpotent matrix", as given in Definition 2.6, fit with each other: a map is nilpotent if and only if it is represented by a nilpotent matrix. (Is it that a transformation is nilpotent if an only if there is a basis such that the map's representation with respect to that basis is a nilpotent matrix, or that any representation is a nilpotent matrix?)

Answer

For any basis  , a transformation   is nilpotent if and only if   is a nilpotent matrix. This is because only the zero matrix represents the zero map and so   is the zero map if and only if   is the zero matrix.

Problem 15

Let   be nilpotent of index four. How big can the rangespace of   be?

Answer

It can be of any size greater than or equal to one. To have a transformation that is nilpotent of index four, whose cube has rangespace of dimension  , take a vector space, a basis for that space, and a transformation that acts on that basis in this way.

 


--possibly other, shorter, strings--

So the dimension of the rangespace of   can be as large as desired. The smallest that it can be is one— there must be at least one string or else the map's index of nilpotency would not be four.

Problem 16

Recall that similar matrices have the same eigenvalues. Show that the converse does not hold.

Answer

These two have only zero for eigenvalues

 

but are not similar (they have different canonical representatives, namely, themselves).

Problem 17

Prove a nilpotent matrix is similar to one that is all zeros except for blocks of super-diagonal ones.

Answer

A simple reordering of the string basis will do. For instance, a map that is assoicated with this string basis

 

is represented with respect to   by this matrix

 

but is represented with respect to   in this way.

 
This exercise is recommended for all readers.
Problem 18

Prove that if a transformation has the same rangespace as nullspace. then the dimension of its domain is even.

Answer

Let   be the transformation. If   then the equation   shows that   is even.

Problem 19

Prove that if two nilpotent matrices commute then their product and sum are also nilpotent.

Answer

For the matrices to be nilpotent they must be square. For them to commute they must be the same size. Thus their product and sum are defined.

Call the matrices   and  . To see that   is nilpotent, multiply  , and  , etc., and, as   is nilpotent, that product is eventually zero.

The sum is similar; use the Binomial Theorem.

Problem 20

Consider the transformation of   given by   where   is an   matrix. Prove that if   is nilpotent then so is  .

Answer

Some experimentation gives the idea for the proof. Expansion of the second power

 

the third power

 

and the fourth power

 

suggest that the expansions follow the Binomial Theorem. Verifying this by induction on the power of   is routine. This answers the question because, where the index of nilpotency of   is  , in the expansion of  

 

for any   at least one of the   and   has a power higher than  , and so the term gives the zero matrix.

Problem 21

Show that if   is nilpotent then   is invertible. Is that "only if" also?

Answer

Use the geometric series:  . If   is the zero matrix then we have a right inverse for  . It is also a left inverse.

This statement is not "only if" since

 

is invertible.